# Differentiation of tan inverse x

Here you will learn differentiation of tan inverse x or arctanx x by using chain rule.

Let’s begin –

## Differentiation of tan inverse x or $$tan^{-1}x$$ :

The differentiation of $$tan^{-1}x$$ with respect to x is $$1\over {1 + x^2}$$.

i.e. $$d\over dx$$ $$tan^{-1}x$$ = $$1\over {1 + x^2}$$.

## Proof using chain rule :

Let y = $$tan^{-1}x$$. Then,

$$tan(tan^{-1}x)$$ = x

$$\implies$$ tan y = x

Differentiating both sides with respect to x, we get

$$d\over dx$$(tan y) = $$d\over dx$$(x)

$$d\over dx$$ (tan y) = 1

By chain rule,

$$sec^2 y$$ $$dy\over dx$$ = 1

$$dy\over dx$$ = $$1\over sec^2 y$$

[ $$\because$$ 1 + $$tan^2 y$$ = $$sec^2 y$$

$$dy\over dx$$ = $$1\over {1 + tan^2 y}$$

$$\implies$$ $$dy\over dx$$ = $$1\over {1 + x^2}$$

$$\implies$$ $$d\over dx$$ $$tan^{-1}x$$ = $$1\over {1 + x^2}$$

Hence, the differentiation of $$tan^{-1}x$$ with respect to x is $$1\over {1 + x^2}$$.

Example : What is the differentiation of $$tan^{-1} x^2$$ with respect to x ?

Solution : Let y = $$tan^{-1} x^2$$

Differentiating both sides with respect to x and using chain rule, we get

$$dy\over dx$$ = $$d\over dx$$ ($$tan^{-1} x^2$$)

$$dy\over dx$$ = $$1\over {1 + x^4}$$.(2x) = $$2x\over {1 + x^4}$$

Hence, $$d\over dx$$ ($$tan^{-1} x^2$$) = $$2x\over {1 + x^4}$$

Example : What is the differentiation of 2x + $$tan^{-1} x$$ with respect to x ?

Solution : Let y = 2x + $$tan^{-1} x$$

Differentiating both sides with respect to x, we get

$$dy\over dx$$ = $$d\over dx$$ (2x) + $$d\over dx$$ ($$tan^{-1} x$$)

$$dy\over dx$$ = 2 + $$1\over {1 + x^2}$$

Hence, $$d\over dx$$ (2x + $$tan^{-1} x$$) = 2 + $$1\over {1 + x^2}$$

### Related Questions

What is the Differentiation of tanx ?

What is the Integration of Tan Inverse x ?

What is the Differentiation of cos inverse x ?